Also, let \({{\nabla {F}}^{\mathsf {P},(k)}}_m\)  denote the \(m\) -th row of \({{\nabla {F}}^{\mathsf {P},(k)}}\) .
Since the consensus error is zero mean [1]} across the ESs, we have \( \mathbf {1}^\top \left({\nabla {F}}^{\mathsf {L},(k)} -{\nabla {F}}^{\mathsf {A},(k)}\right)=\mathbf {0}\)  implying that \((\mathbf {1} \mathbf {1}^\top )\left({\nabla {F}}^{\mathsf {L},(k)} -{\nabla {F}}^{\mathsf {A},(k)}\right)=\mathbf {0}\) . Thus, we get
\(\begin{aligned}{\nabla {F}}^{\mathsf {L},(k)} -{{\nabla {F}}^{\mathsf {P},(k)}}&=\left(\mathbf {I}-\frac{\mathbf {1} \mathbf {1}^\top }{M}\right)\left({\nabla {F}}^{\mathsf {L},(k)} -{{\nabla {F}}^{\mathsf {P},(k)}}\right)=\left(\mathbf {I}-\frac{\mathbf {1} \mathbf {1}^\top }{M}\right)\left( \left(\mathbf {\Lambda }^{(k)}\right)^{\varphi ^{(k)}}{\nabla {F}}^{\mathsf {A},(k)}-{{\nabla {F}}^{\mathsf {P},(k)}}\right)\\&=\left(\mathbf {I}-\frac{\mathbf {1} \mathbf {1}^\top }{M}\right)\left(\left(\mathbf {\Lambda }^{(k)}\right)^{\varphi ^{(k)}} {\nabla {F}}^{\mathsf {A},(k)}-\left(\mathbf {\Lambda }^{(k)}\right)^{\varphi ^{(k)}}{{\nabla {F}}^{\mathsf {P},(k)}}\right)\\&=\left(\left(\mathbf {\Lambda }^{(k)}\right)^{\varphi ^{(k)}}-\frac{\mathbf {1} \mathbf {1}^\top }{M}\right)\left({\nabla {F}}^{\mathsf {A},(k)}-{{\nabla {F}}^{\mathsf {P},(k)}}\right),\end{aligned}\) 
